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If I have a shape, no matter how many time I rotate and translate it with whatever order. I can use a rotation AND THEN a translation to put the shape exactly at the location that I want with the correct orientation.

Does it right?

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Yes. You that's right. –  Hauke Strasdat Oct 9 '11 at 16:31
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2 Answers

This is true in any dimension Euclidean space, but it happens to be particularly easy to see in 2 dimensions.

Rotations and Translations are called isometries as they maintain distances and angles (and pretty much everything). They are special in that they maintain orientation, too. There are 2 more in 2 dimensions: reflections and glide reflections (a combination of a reflection and a translation). These two do not maintain orientation.

So one can see the truth of your statement by noting that any combination or rotations and translations is still an orientation-preserving isometry. So if you translate the object back to its original position (in particular, so that 1 point is where it initially started), then we may take that point as the origin in both the new and initial configurations of our shape. Since everything has been an isometry, our shape is just a rotation away from getting back to its initial state (Rotations around a point form a group - they have inverses). Finally, doing a rotation and then a translation is the same as translating and then rotating (not true in general with isometries - they do not always commute like this).

If that's all too technical, the main idea is that a single rotation can 'orient' our shape in the correct way, and a translation will take it back to the original position always, as the shape has never deformed.

So yes, your statement is true.

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So, when you say rotation here, does it only hold for rotate in the object center? –  zmarcoz Oct 9 '11 at 20:38
    
"Finally, doing a rotation and then a translation is the same as translating and then rotating." That's not true, or at least, not with the same translation and rotation both times. –  Rahul Oct 9 '11 at 20:42
    
Thanks Rahul, but I am asking would the shape to be total match if the sequence is difference and the values (translation and rotation) are allow to be difference. –  zmarcoz Oct 9 '11 at 20:56
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Given a composition $f:=f_n\circ\ldots\circ f_1$ of translations and rotations (about arbitrary centers) of ${\mathbb R}^2$ the resulting map: $f:\ {\mathbb R}^2\to {\mathbb R}^2$ is certainly an isometry which preserves the orientation. Now it is a theorem (which needs a proof) that any such map $f$ can in fact be written as a product of only two factors, namely $\ T\circ R$ (or $R\circ T$, if preferred), where $T$ is a translation and $R$ is a rotation about the origin. In the three-dimensional case replace "centers" by "axes" and "origin" by "axis through the origin".

For the proof write each $f_k$ as $f_k: z\mapsto R_k z+ c_k$ where $R_k$ is a rotation about $O$. Then the rotational part $R$ of the composition $f$ is nothing but $R_n\circ\ldots\circ R_1$, and $f$ differs from $R$ by a translation $T$.

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Do you have the answer for that? I am an engineer, do not have strong MATH background. –  zmarcoz Oct 9 '11 at 20:36
    
@zmarcoz: What do you mean by "the answer"? If you want more detail you have to tell us how your rotations and translations are encoded (as matrices, giving center and angle of rotation, as linear maps of the complex plane, or $\ldots$). –  Christian Blatter Oct 10 '11 at 7:55
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